Ganita Manjari Class 9 Chapter 2 Solutions Introduction to Linear Polynomials

By using Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 2 Introduction to Linear Polynomials NCERT Solutions, students can improve their problem-solving skills.

Introduction to Linear Polynomials Class 9 Solutions

Class 9 Ganita Manjari Chapter 2 Solutions

Class 9 Maths Ganita Manjari Chapter 2 Solutions Introduction to Linear Polynomials

Think and Reflect (NCERT Textbook Page No. 17)

Question 1.
Can you identify the terms, variables and coefficients of this algebraic expression?
Solution:
Given algebraic expression = 200l + 160w + 50lw
Terms: 200l, 160w and 50lw
Variables: l and w
Coefficients: 200 is of variable l, 160 is of variable w, and 50 is of variable lw.

Question 2.
How is it different from the algebraic expression in Example 1?
Solution:
The algebraic expression given in example is 4x + 5y + 3.
It is a linear polynomial in variables x and y.
While the algebraic expression given in Example 2 is 200l + 160w + 50lw, which has two variables, l and w, and the expression is not a linear polynomial because it contains a product term of two variables, lw.

Think and Reflect (NCERT Textbook Page No. 17)

Question 1.
Can you identify the terms, variables and coefficients of this algebraic expression?
Solution:
Yes, the given algebraic expression is 10x – x²
Here, Terms: 10x and -x²
Variables: x
Coefficients: 10 of x and -1 of -x²

Question 2.
Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?
Solution:
Yes.
Similarity: Both expressions use the same variable (x).
Difference: The algebraic expression given in Example 1 is 4x + 5y + 3, which is a linear polynomial with variables x and y.
While the algebraic expression given in Example 3 is 10x – x², which is a quadratic expression with variable x, because it contains the term x × x.

Think and Reflect (NCERT Textbook Page No. 19)

Question 1.
Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm, and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?
Solution:
We know, perimeter of a square of side a cm = 4a cm
∴ Perimeter of square of side 1 cm = 4 × 1 = 4 cm
Perimeter of a square of side 1.5 cm = 4 × 1.5 = 6 cm
Perimeter of a square of side 2 cm = 4 × 2 = 8 cm
Perimeter of a square of side 2.5 cm = 4 × 2.5 = 10 cm
Perimeter of a square of side 3 cm = 4 × 3 = 12 cm
Observation: Every time the side increases by 0.5 cm, the perimeter increases by exactly 2 cm.
This constant rate of change a hallmark of linear polynomials.

Ex 2.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.1 Solutions

Exercise 2.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.1 Solutions

Question 1.
Find the degrees of the following polynomials:
(i) 2x² – 5x + 3
(ii) y³ + 2y – 1
(iii) -9
(iv) 4z – 3
Solution:
We know that the highest power of the variable in a polynomial is called its degree.
(i) In the polynomial, the highest power of x is 2.
So, its degree is 2.
(ii) In the polynomial, the highest power of y is 3.
So, its degree is 3.
(iii) It is a constant polynomial; the highest power of the variable is 0.
So, its degree is 0.
(iv) In the polynomial, the highest power of z is 1.
So, its degree is 1.

Question 2.
Write polynomials of degrees 1, 2, and 3.
Solution:
A polynomial in x of degree 1 is 2x + 7.
A polynomial in x of degree 2 is x² + 3x + 1.
A polynomial in x of degree 3 is x³ + 1.

Question 3.
What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?
Solution:
The polynomial is x⁴ – 3x³ + 6x² – 2x + 7.
Clearly, the coefficient of x² is 6, and the coefficient of x³ is -3.

Question 4.
What is the coefficient of z in the polynomial 4z³ + 5z² – 11?
Solution:
The given polynomial is 4z³ + 5z² – 11.
There is no explicit z term.
Hence, the coefficient of z is 0.

Question 5.
What is the constant term of the polynomial 9x³ + 5x² – 8x – 10?
Solution:
We know the constant term is the term with no variable.
In 9x³ + 5x² – 8x – 10, the constant term is -10.

Ex 2.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.2 Solutions

Exercise 2.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.2 Solutions

Question 1.
Find the value of the linear polynomial 5x – 3 if:
(i) x = 0
(ii) x = -1
(iii) x = 2
Solution:
(1) The value of 5x – 3 for x = 0 is
5(0) – 3 = 0 – 3 = -3
(ii) The value of 5x – 3 for x = -1 is
5(-1) – 3 = -5 – 3 = -8
(iii) The value of 5x – 3 for x = 2 is
5(2) – 3 = 10 – 3 = 7

Question 2.
Find the value of the quadratic polynomial 7s² – 4s + 6 if:
(i) s = 0
(ii) s = -3
(iii) s = 4
Solution:
(i) The value of 7s² – 4s + 6 for s = 0 is 7(0)² – 4(0) + 6 = 0 – 0 + 6 = 6
(ii) The value of 7s² – 4s + 6 for s = -3 is 7(-3)² – 4(-3) + 6 = 63 + 12 + 6 = 81
(iii) The value of 7s² – 4s + 6 for s = 4 is 7(4)² – 4(4) + 6 = 112 – 16 + 6 = 102

Question 3.
The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages.
Solution:
Let Salil’s present age be x years.
Then, his mother’s present age is 3x years.
Salil’s age (in years) after 5 years will be x + 5; and
His mother’s age (in years) after 5 years will be 3x + 5
Now, after 5 years, their ages add up to 70 years
∴ (x + 5) + (3x + 5) = 70
⇒ 4x + 10 = 70
⇒ 4x = 60
⇒ x = 15
Hence, Salil’s present age is 15 years, and his mother’s present age is 45 years.
Verification:
After 5 years, 20 + 50 = 70

Question 4.
The difference between two positive integers is 63. The ratio of the two integers is 2 : 5. Find the two integers.
Solution:
Since the two positive integers are in the ratio 2 : 5, we can take them as 2x and 5x.
Now the difference between the two positive integers is 63.
∴ 5x – 2x = 63
⇒ 3x = 63
⇒ x = 21
Thus, the two positive integers are 42 (21 × 2) and 105 (21 × 5).
Verification: 105 – 42 = 63
and Ratio = 42 : 105 = 2 : 5

Question 5.
Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total of ₹ 88, how many coins does she have of each type?
Solution:
Let the number of 5-rupee coins be x.
Then, the number of 2-rupee coins is 3x.
Now, as she has a total of ₹ 88.
∴ 5x + 2(3x) = 88
⇒ 11x = 88
⇒ x = 8
Thus, Rubi has 8 coins of 5-rupee and 24 (3 × 8) coins of 2-rupee.
Verification:
5(8) + 2(24) = 40 + 48 = 88

Question 6.
A farmer cuts a 300-foot fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?
Solution:
Let the length (in feet) of the shorter piece be x.
Then, the length (in feet) of the longer piece is 300 – x.
Now, the longer piece is 4 times the shorter one
∴ 300 – x = 4x
⇒ 5x = 300
⇒ x = 60
Thus, the lengths of the two pieces are 60 feet and 240 feet.
Verification: 60 + 240 = 300

Question 7.
If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?
Solution:
Let the width (in cm) of the rectangle be x.
Then, the length (in cm) of the rectangle is 3 + 2x.
Perimeter of the rectangle = 2(length + width)
= 2[(3 + 2x) + x]
= 2(3 + 3x)
= 6 + 6x
Now the perimeter is 24 cm
∴ 6x + 6 = 24
⇒ 6x = 18
⇒ x = 3
Thus, the dimensions of the rectangle are 9 cm × 3 cm.
Verification:
Perimeter = 2(9 + 3)
= 2 × 12
= 24 cm

Ex 2.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.3 Solutions

Exercise 2.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.3 Solutions

Solve the following:

Question 1.
A student has ₹ 500 in her savings bank account. She gets ₹ 150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nth month.
Solution:
Given, initial amount in savings bank Account = ₹ 500
and monthly pocket money = ₹ 150
Amount in the Account at the end of first month = ₹ 500 + ₹ 150 = ₹ 650
Amount in the Account at the end of second month = ₹ [500 + 2(150)] = ₹ 800
Amount in the Account at the end of third month = ₹ [500 + 3(150)] = ₹ 950
Amount in the Account at the end of fourth month = ₹ [500 + 4(150)] = ₹ 1100
Amount in the Account at the end of fifth month = ₹ [500 + 5(150)] = ₹ 1250
Similarly, amount in the Account at the end of nth month = ₹ [500 + n(150)] = 500 + 150n
Hence, the linear expression to represent the amount she will have in the nth month is A_n = 150n + 500

Question 2.
A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3,…. hours? Find a linear expression to represent the number of members at the end of the nth hour.
Solution:
Given, initial number of members in the rally = 120
and number of dropouts every hour = 9
So, after 1 hour, the number of members in the rally = 120 – 1(9) = 111
After 2 hours, the number of members in the rally = 120 – 2(9) = 102
After 3 hours, the number of members in the rally = 120 – 3(9) = 93, and so on.
Hence, the linear expression to represent the number of members in the rally at the end of the nth hour is A_n = 120 – n(9) = 120 – 9n.

Question 3.
Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle.
Solution:
We know area of rectangle = length × breadth = l × b
Now, length (in cm) of a rectangle is 13.
(i) When breadth (in cm) of the rectangle is 12,
the area of the rectangle is = (13 × 12) sq cm = 156 sq cm.
(ii) When breadth (in cm) of the rectangle is 10,
the area of the rectangle is = (13 × 10) sq cm = 130 sq cm.
(iii) When breadth (in cm) of the rectangle is 8,
the area of the rectangle is = (13 × 8) sq cm = 104 sq cm
Hence, the linear expression representing the area of the rectangle is A_n = 13 × b = 13b cm²
where b is the breadth of the rectangle.

Question 4.
Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the rectangular box.
Solution:
Length and breadth (in cm) of a rectangular box are 7 and 11.
We know that the volume of a rectangular box = length × breadth × height = l × b × h
(i) When height (in cm) of the rectangular box is 5,
the volume of the rectangular box is (7 × 11 × 5) cu cm = (77 × 5) cu cm = 385 cu cm.
(ii) When the height (in cm) of the rectangular box is 9,
the volume of the rectangular box is (7 × 11 × 9) cu cm = (77 × 9) cu cm = 693 cu cm.
(iii) When height (in cm) of the rectangular box is 13,
the volume of the rectangular box is (7 × 11 × 13) cu cm = (77 × 13) cu cm = 1001 cu cm.
Hence, the linear expression representing the volume of the rectangular box is A_n = (7 × 11 × h) cu cm = 77 h cu cm
where h is the height of the rectangular box.

Question 5.
Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.
Solution:
Initial number of pages in a book = 500
The number of pages Sarita reads every day = 20
So, in 15 days, the number of pages she will be able to read = 20 × 15 = 300
Hence, the number of pages left to be read after 15 days = 500 – 300 = 200
Hence, the linear expression representing the number of pages left to be read after n days is A_n = 500 – 200n

Ex 2.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.4 Solutions

Exercise 2.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.4 Solutions

Question 1.
Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.
(i) Find the height after 7 months.
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
Solution:
It is given that
Initial height (in feet) of the plant = 1.75
Monthly increase in height (in feet) = 0.5
So, (i) height of the plant after 7 months = 1.75 + 7(0.5)
= 1.75 + 3.5
= 5.25 feet
(ii)


(iii) h = 1.75 + 0.5t
It represents linear growth as the height of the plant increases every month by the same height of 0.5 feet.

Question 2.
A mobile phone is bought for ₹ 10,000. Its value decreases by ₹ 800 every year.
(i) Find the value of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
Solution:
It is given that
Initial price (in ₹) of the mobile phone = ₹ 10,000
Yearly decrease in the value (in ₹) = ₹ 800
So, (i) Value of the mobile phone after 3 years = ₹ 10,000 – 3(₹ 800)
= ₹ 10,000 – ₹ 2400
= ₹ 7600
(ii)

(iii) Expression: v = 10,000 – 800t
It represents linear decay as the value of the mobile phone decreases every year by the same amount of ₹ 800.

Question 3.
The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find the population of the village after 6 years.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
(iii) Find an expression that relates P and t, and explain why it represents linear growth.
Solution:
It is given that
Initial population of a village = 750
Yearly movement of population from nearby city to the village = 50
So, (i) population of the village after 6 years = 750 + 6(50)
= 750 + 300
= 1050
(ii)

(iii) Expression: P = 750 + 50t
It represents linear growth as the population of the village increases every year by the same number, 50.

Question 4.
A telecom company charges ₹ 600 for a certain recharge scheme. This prepaid balance is reduced by ₹ 15 each day after the recharge.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
Solution:
It is given that
Initial amount of recharge = ₹ 600
Per day reduction in prepaid balance = ₹ 15
So, (i) pre-paid balance after x days = ₹ [600 – x(15)] = ₹ [600 – 15x]
It represents linear decay as the pre-paid balance decreases every day by the same amount, ₹ 15.
(ii) Pre-paid balance after x days = 600 – 15x
The balance will run out when 600 – 15x becomes 0.
⇒ 600 – 15x = 0
⇒ 600 = 15x
⇒ x = 40
Thus, the balance will run out after 40 days.
(iii) b(x) = 600 – 15x

Ex 2.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.5 Solutions

Exercise 2.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.5 Solutions

Question 1.
A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹ 400. When she accessed 14 modules, her bill was ₹ 500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.
Solution:
Here, no. of modules = x
monthly bill = y
When x = 10, y = 400
When x = 14, y = 500
Using the relation y = ax + b, we have
400 = 10a + b …..(i)
and 500 = 14a + b ……(ii)
Subtracting (i) from (ii), we get
500 – 400 = 14a – 10a + b – b
⇒ 100 = 4a
⇒ a = 25
Putting a = 25 in (i), we have
400 = 10(25) + b
⇒ b = 400 – 250
⇒ b = 150
Thus, a = 25 and b = 150.

Question 2.
A gym charges a fixed monthly fee plus an additional hourly fee for using the badminton court. A student at the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹ 1100. If the monthly bill y depends on the hours of use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.
Solution:
Here, when x = 10, y = 800
When x = 15, y = 1100
Using the relation y = ax + b, we have
800 = 10a + b ……(i)
and 1100 = 15a + b …..(ii)
Subtracting eq. (i) from eq. (ii), we get
1100 – 800 = 15a – 10a + b – b
⇒ 300 = 5a
⇒ a = 60
Putting a = 60 in (i), we have
800 = 10(60) + b
⇒ 800 – 600 = b
⇒ 200 = b
⇒ b = 200
Thus, a = 60 and b = 200.

Question 3.
Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.
(Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)
Solution:
Here, When, °C = 0, °F = 32
When, °C = 100, °F = 212
Using the relation °C = a °F + b, we have
0 = a(32) + b …..(i)
and 100 = a(212) + b ……(ii)
Subtracting (i) from (ii), we get
⇒ 100 – 0 = 212a – 32a
⇒ 100 = 180a
⇒ a = \(\frac{100}{180}=\frac{5}{9}\)
Putting a = \(\frac {5}{9}\) in (ii), we have
100 = 212(\(\frac {5}{9}\)) + b
⇒ 100 = \(\frac{212 \times 5}{9}\) + b
⇒ 100 – \(\frac {1060}{9}\) = b
⇒ \(\frac{900-1060}{9}\) = b
⇒ \(-\frac {160}{9}\) = b
or b = \(-\frac {160}{9}\)
Thus, a = \(\frac {5}{9}\) and b = \(-\frac {160}{9}\)
Thus, the linear relationship between °C and °F is:
\({ }^{\circ} \mathrm{C}=\left(\frac{5}{9}\right){ }^{\circ} \mathrm{F}-\frac{160}{9} \text { or }{ }^{\circ} \mathrm{C}=\left(\frac{5}{9}\right)\left[{ }^{\circ} \mathrm{F}-32\right]\)

Ex 2.6 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.6 Solutions

Exercise 2.6 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 2.6 Solutions

Question 1.
Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.
(i) y = 4x, y = 2x, y = x
(ii) y = -6x, y = -3x, y = -x
(iii) y = 5x, y = -5x
(iv) y = 3x – 1, y = 3x, y = 3x + 1
(v) y = -2x – 3, y = -2x, y = 2x + 3
Solution:
(i) Given lines are
y = 4x ….(i)
y = 2x ……(ii)
y = x ……(iii)
Putting x = 0 in eq. (i), we get
y = 4 × 0 = 0
It passes through the origin (0, 0).
Similarly, putting x = 0 in eq. (ii), we get
y = 2 × 0 = 0
⇒ y = 0
It also passes through the origin (0, 0).
Putting x = 0 in eq. (iii), we get
y = 0
Hence, all three lines pass through the origin.
Now to draw the line y = 4x, we just need two points that satisfy the equation.
Let’s pick values for x and calculate y.
Now line (i), y = 4x
When x = 1, y = 4 × (1) = 4
When x = 2, y = 4 × (2) = 8
Table of values

Now mark a point O on the graph paper as the origin.
Draw two lines X’X and YY’ perpendicular to each other and passing through the origin O.
Scale: take 1 cm = 1 unit on x-axis
1 cm = 1 unit on the y-axis
To plot the point (1, 4), start from the origin O, move along OX and reach the point marked 1.
From here, move 4 units upwards parallel to the y-axis.
Mark the point A and write (1, 4).
To plot point (2, 8), start from the origin O, move along OX, and reach the point marked 2.
From here, move 8 units upwards parallel to the y-axis.
Mark the point B and write (2, 8) there.
Join these points to form a line AB.

Now line (ii), y = 2x
When x = 1, y = 2 × 1 = 2
When x = 2, y = 2 × 2 = 4
Table of values

Now we plot the points C(1, 2) and D(2, 4) and join these points by a ruler to form the line CD representing eq. (ii) as shown in the graph.

and line (iii), y = x
When x = 1, y = 1
When x = 2, y = 2
Table of values

Similarly we plot the points E(1, 1) and F(2, 2).
Now join these points to form the line EF representing eq. (iii) as shown in the graph.

Reflect on the role of a and b
(i) Since b = 0, all lines pass through the origin.
(ii) All lines have positive slope (a > 0)
So they are increasing lines.
(iii) The greater the value of a, the steeper the line.
These equations represent lines in the form y = ax (b = 0), where a represents the slope of the line.

(ii) Given lines are
y = -6x ……(i)
y = -3x ……(ii)
y = -x ……(iii)
Putting x = 0 in eq. (i), we get y = 0
∴ Eq. (i) passes through the origin.
Putting x = 0 in eq. (ii), we get y = 0
∴ Eq. (ii) passes through the origin.
Again, putting x = 0 in eq. (iii), we get y = 0.
Hence, all three lines pass through the origin.
Now to draw the line y = -6x, we need two points that satisfy the equation.
Let’s pick values for x and calculate y.
Here, (i ) When x = 1, y = -6(1) = -6
When x = 2, y = -6(2) = -12
Table of values

Now mark a point O on the graph paper as the origin.
Draw a horizontal line OX (x-axis) and a vertical line OY (y-axis) through O.
Scale: take 1 cm = 1 unit on x-axis
1 cm = 1 unit on the y-axis
To plot the point (1, -6), start from the origin O, move along OX and reach the point marked 1.
From here, move -6 units downwards parallel to the y-axis.
Mark the point A and write A(1, -6).
Again, to plot the point (2, -12), start from the origin, move along OX and reach the point marked 2.
From here move -12 units downwards parallel to the y-axis and mark the point B(2, -12).
Join these points to form a line AB.

Now (ii), y = -3x
When x = 1, y = -3
When x = 2, y = -6
Table of values

To plot the point (1, -3) start from the origin O, move along OX and reach the point marked 1.
From here move -3 units downwards parallel to the y-axis.
Mark the point C and write C(1, -3).
Again, to plot point (2, -6), start from the origin 0 move along OX, and reach the point marked 2.
From here, move -6 units downwards parallel to the y-axis.
Mark the point D(2, -6).
Join these points to form a line CD.

Now (iii), y = -x
When x = 1, y = -1
When x = 3, y = -3
Table of values

Similarly, we plot the points E(1, -1) and F(3, -3) on the above graph paper and join these points by a ruler to form the line EF representing eq. (iii) as shown in the graph.

Reflect on the role of a and b
All lines pass through the origin.
Negative ‘a’ means lines slope downward.
Larger magnitude of ‘a’
⇒ steeper downward slope.

(iii) The given lines are
y = 5x …….(i)
y = -5x …..(ii)
From (i), y = 5x
When x = 1, y = 5
and When x = 2, y = 10
Thus, we have the following table of values:
Table of Values

Plotting the points A(1, 5) and B(2, 10) and joining them to get a straight line.
The straight line obtained is the graph of y = 5x.

From (ii), y = -5x
When, x = 1 then y = -5 and
When x = -1, then y = 5
Thus we have the following table of values:
Table of Values

Plotting the points C(1, -5) and D(-1, 5) and joining them to get a straight line.
The straight line thus obtained is the graph of y = -5x.

Reflect on the role of a and b
(i) These equations represent lines in the form y = ax (b = 0); a represents the slope of the line.
(ii) Same magnitude of ‘a’ → same steepness but in opposite direction.
(iii) y = 5x steeps upwards and y = -5x steeps downwards.
(iv) Both lines pass through the origin.

(iv) We have,
y = 3x – 1 …….(i)
y = 3x ……(ii)
y = 3x + 1 …….(iii)
From (i), y = 3x – 1
When x = 1, then y = 3 × 1 – 1 = 2
When x = 2, then y = 3 × 2 – 1 = 5
Thus, we have the following table of values:
Table of Values

Now, plotting the points A(1, 2) and B(2, 5) and joining them to get a straight line.
The straight line thus obtained is the graph of y = 3x – 1.

From (ii), y = 3x
When x = 1, y = 3 × 1 = 3
When x = 2, y = 3 × 2 = 6
Thus, we have the following table of values:
Table of Values

Plotting the points C(1, 3) and D(2, 6) and joining them to get a straight line.
The straight line thus obtained is the graph of y = 3x.

Now from (iii), y = 3x + 1
When x = 1, y = 3 × 1 + 1 = 4
When x = 2, y = 3 × 2 + 1 = 7
Thus, we have the following table of values.
Table of Values

Now, plot the points E(1, 4) and F(2, 7) and join them to get a straight line.
The straight line thus obtained is the graph of y = 3x + 1.

Reflect on the role of a and b
(i) These equations represent lines in the form y = ax + b; a represents the slope of the line and b represents the y-intercept.
(ii) The lines are parallel as their slope (a is the same (3).
b = -1, b = 0 and b = 1 indicate that the line is below, passes through and above the origin (0, 0).

(v) Given lines are
y = -2x – 3 …..(i)
y = -2x …..(ii)
y = 2x + 3 ……(iii)
From (i),
When x = -1, y = -2 × (-1) – 3 = -1
When x = 0, y = -2 × 0 – 3 = -3
Thus, we have the following table of values:
Tables of Values

Plotting the points A(-1, -1) and B(0, -3) and joining them to get a straight line.
The straight line thus obtained is the graph of y = -2x – 3.

Again from (ii), y = -2x
When x = 1, y = -2
When x = 2, y = -2 × 2 = 4
Thus, we have the following table of values:
Tables of Values

Plotting the points C(1, -2) and D(2, -4) and joining them to get a straight line.
The straight line thus obtained is the graph of y = -2x.

Again from (iii), y = 2x + 3
When, x = 1, y = 2 × 1 + 3 = 5
When x = 2, y = 2 × 2 + 3 = 7
Thus, we have the following table of values.
Tables of Values

Plotting the points E(1, 5) and F(2, 7) and joining them to get a straight line.
The straight line thus obtained is the graph of y = 2x + 3.

Reflect on the role of a and b
(i) These equations represent lines in the form y = ax + b; a represents the slope of the line and b represents the y-intercept.
(ii) The lines y = -2x – 3 and y = -2x are parallel as their slope (a) is -2.
(iii) Line y = 2x + 3 has +ve slope – different direction.
(vi) b = -3, b = 0 and b = 3 indicate that the line is below, passes through and above the origin(0, 0).

Ganita Manjari Class 9 Maths Chapter 2 End of Chapter Exercise Solutions

Introduction to Linear Polynomials End of Chapter Exercise Solutions

Question 1.
Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is -7.
Solution:
Here is a simple example of a polynomial of degree 3 in the variable x, with the coefficient of the x² term equal to -7.
P(x) = 2x³ – 7x² + 5x – 1
The degree is 3 because the highest power of x is x³.
The coefficient of x² is -7 as required.
The other coefficients (2, 5, -1) can be chosen freely.
Since the problem only specifies the condition on the x² term.

Question 2.
Find the values of the following polynomials at the indicated values of the variables:
(i) 5x² – 3x + 7 if x = 1
(ii) 4t³ – t² + 6 if t = a
Solution:
(i) The value of the polynomial 5x² – 3x + 7 if x = 1 is
5(1)² – 3(1) + 7 = 5 – 3 + 7 = 12 – 3 = 9
(ii) The value of the polynomial 4t³ – t² + 6 if t = a is
4(a)³ – (a)² + 6 = 4a³ – a² + 6

Question 3.
If we multiply a number by \(\frac {5}{2}\) and add \(\frac {2}{3}\) to the product, we get \(\frac {-7}{12}\). Find the number.
Solution:
Let the number be x. Then,

Now, x = \(\frac{-5}{2 \times 5}=\frac{-1}{2}\)
Thus, the number is \(-\frac {1}{2}\).

Question 4.
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution:
Let a positive number be x.
Then, the larger number is 5x.
As per the question,
5x + 21 = 2(x + 21)
⇒ 5x – 2x = 42 – 21
⇒ 3x = 21
⇒ x = 7
Thus, the numbers are 7 and 5 × 7 = 35.

Question 5.
If you have ₹ 800 and you save ₹ 250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.
Solution:
Here initial amount = ₹ 800
Monthly savings = ₹ 250
So, the linear pattern after n months is = 800 + 250n
(i) After 6 months, you will have
800 + 250 × 6 = ₹ 2300
(ii) After 2 years (24 months), you will have
800 + 250 × 24 = 800 + 6000 = ₹ 6800

Question 6.
The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.
Solution:
Let the tens digit = x and the units digit = y.
So the original number is 10x + y
The interchanged number is 10y + x
Digits differ by 3
x – y = 3 or y – x = 3
Sum of original and interchanged = 143
(10x + y) + (10y + x) = 143
Simplify:
11x + 11y = 143
⇒ x + y = 13
We now have:
x + y = 13 …(i)
x – y = 3 …(ii)
(taking the positive difference case)
Add these equations:
(x + y) + (x – y) = 13 + 3
⇒ 2x = 16
⇒ x = 8
Substitute into x + y = 13:
8 + y = 13
⇒ y = 5
So the original number is
10x + y = 10 × 8 + 5 = 80 + 5 = 85
and the interchanged number is
10y + x = 10 × 5 + 8 = 58.
Verification:
85 + 58 = 143
Digits differ by 3 (8 – 5 = 3)
Hence, the two members are 85 and 58.

Question 7.
Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
(i) y = -3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?
Solution:
(i) The given line is y = -3x + 4
When, x = 0, y = -3 × 0 + 4 = 4
When, x = 1, y = -3 × 1 + 4 = 1
Thus, we have the following table of values.
Table of Values

Plotting the points A(0, 4) and B(1, 1) and joining them to get a straight line.
The straight line thus obtained is the graph of y = -3x + 4.
Slope is -3, y-intercept is 4.
The line cuts the y-axis at the point (0, 4).

(ii) Given line is 2y = 4x + 7
⇒ y = 2x + \(\frac {7}{2}\)
Now, When, x = 0, y = \(\frac {7}{2}\)
When, x = 2, y = 2 × 2 + \(\frac {7}{2}\)
= 4 + \(\frac {7}{2}\)
= \(\frac {15}{2}\)
Thus, we have the following table of values.
Table of Values

Plotting the points A (0, \(\frac {7}{2}\)) and B(2, \(\frac {15}{2}\)) and join them to get a straight line.
The straight line thus obtained is the graph of 2y = 4x + 7.

Slope is 2, y-intercept is \(\frac {7}{2}\)
The line cuts the y-axis at the point (0, \(\frac {7}{2}\)).

(iii) We have 5y = 6x – 10
⇒ y = \(\frac {6x}{5}\) – 2
When x = 0, y = \(\frac{6 \times 0}{5}\) – 2 = -2
When x = 2, y = \(\frac{6 \times 2}{5}\) – 2 = \(\frac {-2}{5}\)
Thus, we have the following table of values.
Table of Values

Plotting the point A(0, -2) and B(2, \(\frac {-2}{5}\)) and join them to get a straight line.
The straight line thus obtained is the graph of 5y = 6x – 10.

Slope is \(\frac {6}{5}\), y-intercept is -2.
The line cuts the y-axis at the point (0, -2).

(iv) We have 3y = 6x – 11
⇒ y = 2x – \(\frac {11}{3}\)
When x = 0, y = 2 × 0 – \(\frac {11}{3}\) = \(-\frac {11}{3}\)
When, x = 2, y = 2 × 2 – \(\frac {11}{3}\) = \(\frac {1}{3}\)
Table of Values

Plotting the points A(0, \(-\frac {11}{3}\)) and B(2, \(\frac {1}{3}\)) and join them to get a straight line.
The straight line thus obtained is the graph of 3y = 6x – 11.

Slope is 2, y-intercept is \(-\frac {11}{3}\)
The line cuts the y-axis at the point (0, \(-\frac {11}{3}\))
Lines are parallel if their slopes are the same.
Here, lines (ii) and (iv) are parallel, as each has slope 2.

Question 8.
If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = \(\frac {9}{5}\)(x – 273) + 32.
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
Solution:
Here, the relation between the two systems Kelvin (K) and Fahrenheit (F), of measurement of temperature is given by
°F = \(\frac {9}{5}\)(°K – 273) + 32
(i) For K = 313, we have
F = \(\frac {9}{5}\)(313 – 273) + 32
= \(\frac {9}{5}\) × 40 + 32
= 72 + 32
= 104

(ii) For F = 158, we have
158 = \(\frac {9}{5}\)(K – 273) + 32 = \(\frac {9}{5}\) × 40 + 32
⇒ \(\frac {9}{5}\)(K – 273) = 126
⇒ K – 273 = 70
⇒ K = 343

Question 9.
The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.
Solution:
We know that “The work done (w) by a body on the application of a constant force (F) is the product of the constant force and the distance (d) travelled by the body in the direction of the force” is expressed as
w = F × d
⇒ w = 3d
Now the equation of the line is w = 3d
When d = 1, w = 3 × 1 = 3
When d = 2, w = 3 × 2 = 6
Thus, we have the following table of values
Table of Values

Plotting the points A(1, 3) and B(2, 6) on graph paper and joining them to get a straight line.
The straight line thus obtained is the graph of w = 3d.
For d = 2, w = 3 × 2 = 6

Question 10.
The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x).
(ii) Find the coordinates where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.
Solution:
(i) Let p(x) = ax + b
Since the line [graph of p(x)] passes through the points (1, 5) and (3, 11), we have
5 = a(1) + b
11 = a(3) + b
Solving the two equations for a and b, we have
a = 3 and b = 2
So, p(x) = 3x + 2

(ii) The graph of p(x) will cut y-axis when x = 0.
y = p(0) = 3(0) + 2 = 2
i.e., the graph of p(x) will cut the y-axis at (0, 2).
The graph of p(x) will cut x-axis when y = p(x) = 0
⇒ 3(x) + 2 = 0
⇒ x = \(-\frac {2}{3}\)
i.e., The graph of p(x) will cut x-axis at (\(-\frac {2}{3}\), 0).

(iii) Plotting the points A(1, 5) and B(3, 11) on the graph paper and joining them to get the straight line.
The straight line thus obtained is the graph of p(x).

Question 11.
Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) p(0) = 5.
(ii) The polynomial p(x) – q(x) cuts the x-axis at (3, 0).
(iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x.
Find the polynomials p(x) and q(x).
Solution:
(i) Here, p(x) = ax + b and q(x) = cx + d
Since p(0) = 5, we have
a(0) + b = 5
⇒ b = 5
Hence, p(x) = ax + 5
(ii) Since p(x) + q(x) = 6x + 4.
We have (ax + 5) + (cx + d) = 6x + 4
⇒ (a + c) x + (5 + d) = 6x + 4
⇒ (a + c) = 6 and (5 + d) = 4
⇒ (a + c) = 6 and (d) = -1
Thus, q(x) = cx – 1
Since the graph of p(x) – q(x) cuts x-axis at (3, 0),
p(3) – q(3) = 0 ……(A)
From (i), p(3) = 3a + 5
From (iii), q(3) = 3c – 1
From (A), we have
(3a + 5) – (3c – 1) = 0
⇒ 3(a – c) + 6 = 0
⇒ a – c = -2
Solving the equations
a + c = 6 and a – c = -2
we have a = 2 and c = 4
Thus, p(x) = 2x + 5 and q(x) = 4x – 1.

Question 12.
Look at the first three stages of a growing pattern of hexagons made using matehsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.

(i) Draw the next two stages of the pattern. How many matehsticks will be required at these stages?
(ii) Complete the following table:

(iii) Find a rule to determine the number of matehsticks required for the nth stage.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.
Solution:
Here, Number of matchsticks used in the 1st stage hexagon = 6
Number of matchsticks used in 2nd stage hexagons = 6 + 5 = 11
Number of matehsticks used in 3rd stage hexagons = 11 + 5 = 16
(i) Number of matehsticks used in 4th stage hexagons = 16 + 5 = 21
Number of matehsticks used in 5th stage hexagons = 21 + 5 = 26

(iii) Number of matchsticks used in growing patterns of hexagons are 6, 11, 16, 21, 26,……
Here, 6 = 5(1) + 1;
11 = 5(2) + 1;
16 = 5(3) + 1;
21 = 5(4) + 1;
26 = 5(5) + 1;
Extending this pattern, we have “Number of matehsticks required for the nth stage is 5n + 1”.
(iv) Numbers of matehsticks required for the 15th stage is = 5(15) + 1 = 76
(v) Let 200 matchsticks be used in the nth stage.
Then, 200 = 5n + 1
⇒ 5n = 199
⇒ n = \(\frac {199}{5}\), which is not a natural number.
Hence, 200 matchsticks cannot form a stage in this pattern.

Question 13.
Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, -1).
(iii) The graph of q(x) is parallel to the graph of p(x).
Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.
Solution:
Here, p(x) = ax + b and q(x) = cx + d
Since the graph of p(x) passes through the points (2, 3) and (6, 11), we have
3 = 2a + b, and 11 = 6a + b
Solving these equations for a and b, we have
a = 2; b = -1
Thus p(x) = 2x – 1
Since the graph of q(x) passes through the point (4, -1)
We have, -1 = 4c + d …..(A)
Since the graph of q(x) is parallel to the graph of p(x), their slopes are equal.
So, a = c
⇒ c = 2
From (A), we get d = -9
Thus, q(x) = 2x – 9
Now, the line given by p(x) = 2x – 1 will meet x-axis,
where 2x – 1 = 0, i.e., where x = \(\frac {1}{2}\)
Thus, the point of meet is (\(\frac {1}{2}\), 0)
Also, the line given by q(x) = 2x – 9 will meet x-axis, where
2x – 9 = 0, i.e., where x = \(\frac {9}{2}\)
Thus, the point of meet is (\(\frac {9}{2}\), 0)

Question 14.
What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Solution:
We can rewrite the given linear function
f(x) = ax + a as
f(x) = a(x + 1) ………(i)
1. Slopes of lines:
As a > 0, the slope of the lines given by (i) is positive, and all lines are going upwards from left to right.
2. y-intercepts of lines:
For getting y-intercepts, we take x = 0 in f(x) = a(x + 1)
So, the y-intercept is (0, a)
As a > 0, the y-intercepts are above the origin (0, 0).
3. x-intercepts of lines:
For getting x-intercepts, we take f(x) = 0 in f(x) = a(x + 1)
So, the x-intercept is (-1, 0).
Thus, the linear function f(x) = ax + a have the following in common:

• All graphs have positive slopes.
• The y-intercepts of all graphs are above the origin (0, 0).
• All graphs pass through the fixed point (-1, 0).